Download (a) x^2 – 2x-13=0 (b) 4x^2 - 4x+3=0 (c) x^2 +12x-64=0 (d) 2x^2

(a) x^2 – 2x-13=0
(b) 4x^2 - 4x+3=0
(c) x^2 +12x-64=0
(d) 2x^2 -3x-5=0
(a) x^2 - 2x = 13
4x^2 - 8x = 52
4x^2 - 8x + 4 = 56
(2x - 2)^2 = 56
2x - 2 = 56 = 214
2(x - 1) = 214
x - 1 = 14
x = 1 + 14
2x - 2 = 56 = -214
2(x - 1) = -214
x - 1 = -14
x = 1 - 14
(b) x^2 - 4x = -3
4x^2 - 16x = -12
4(x^2 - 4x) = -12
x^2 - 4x = -3
x^2 - 4x + 4 = 1
(x - 2)^2 = 1
x-2=1
x = 2 + 1 = 2 + 1 = 3
x - 2 = -1
x = 2 - 1 = 2 - 1 = 1
(c) x^2 + 12x = 64
4x^2 + 48x = 256
4x^2 + 48x + 144 = 400
(2x + 12)^2 = 400
2x + 12 = 400
2x + 12 = 20
x=4
2x + 12 = -20
x = -16
(d) 2x^2 - 3x = 5
x^2 - 1.5x = 2.5
4x^2 - 6x = 10
4x^2 - 6x + 2.25 = 12.25
(2x - 1.5)^2 = 12.25
2x - 1.5 = 12.25
2x - 1.5 = 3.5
x = 2.5
2x - 1.5 = -3.5
x = -1
Mathematicians have been searching for a formula that yields prime numbers. One such formula was x^2 - x +41.
Select some numbers for x, substitute them in the formula, and see if prime numbers occur. Try to find a number for
x that when substituted in the formula yields a composite number.
Let p(x) = x^2 - x + 41
Lets plug in x = 1, 2, 3, 5, 7, 10, 12 and 20 and see if we get prime numbers
p(1) = 1^2 - 1 + 41 = 41 (Prime)
p(2) = 2^2 - 2 + 41 = 43 (Prime)
p(3) = 3^2 - 3 + 41 = 47 (Prime)
p(5) = 5^2 - 5 + 41 = 61 (Prime)
p(7) = 7^2 - 7 + 41 = 83 (Prime)
p(10) = 10^2 - 10 + 41 = 131 (Prime)
p(12) = 12^2 - 12 + 41 = 173 (Prime)
p(20) = 20^2 - 20 + 41 = 421 (Prime)
... The list goes on, yielding prime numbers up to x = 40
Now, let x = 41. Then
p(41) = 41^2 - 41 + 41 = 41^2 = 1681, which obviously is not a prime.